GREPhysics.NET: GR9677 #61

GR9677 #61

Problem

GREPhysics.NET Official Solution

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Electromagnetism $\Rightarrow$ }Gauss Law

Recall Gauss Law $\vec{E} \cdot d\vec{A} = q/\epsilon_0$ . Thus, $E(4\pi r^2)=\int_0^{R/2}A r^2 4\pi r^2 dr = 4\pi A (R/2)^5/5$ . Solving for E, one has $E=A/5(R/2)^3$ , as in choice (B).

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Comments

rabia
2012-10-11 01:04:05

I have to solve a problem by using Coulumn' law given rho=rho not (1-r/R) and find Q from given expression.

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physicsphysics
2011-10-11 07:22:32

Does anybody have a problem by using Coulumn' law?
I used $\vec{E}=\frac{1}{4\Pi\e0}\int_0^\R/2\frac{Ar^2}{r^2}r^2Sin\Theta dr d\Theta d\Phi$ . The answer is $\frac{AT^4}{24e0}$ . Can anybody explain what I am wrong?

3Danyon
2011-10-13 08:44:10

I made the same mistake initially. The thing is, one has to calculate the charge first and then divide by the $(R/2)^2$ . Otherwise you change the integrand and get the wrong numerical factor. To spell it out, the field is given by:

$\vec{E}=\frac{1}{4\pi \epsilon_0} \frac{Q}{(R/2)^2} = \frac{1}{4\pi \epsilon_0 (R/2)^2} \cdot \int (A r^2) \cdot r^2 \sin{\theta} dr d\theta d\phi$

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phoxdie
2010-11-10 15:20:06

Sorry to get hung up on this, but it was very confusing the first time I looked at this solution. There is a small "typo" in the solution. A factor of epsilon is dropped for no reason.

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Kabuto Yakushi
2010-09-19 10:28:26

Another method for solving this problem would be to take the volume integral of $\rho$ to solve for Q. Then to plug Q into Gauss' law.

$Q=\int_0^{2\pi}\int_0^\pi\int_0^{\frac{R}{2}}{\rho}d\tau$

where $d\tau=r^2sin(\theta)drd{\theta}d{\phi}$

plug Q into Gauss' law:

$E=\frac{Q}{{\pi}r^2{\epsilon_0}}$

solving one gets choice B).

Yosun's solution is faster, but if using the shell method for integration doesn't come to mind, one must resort to the long method.

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tsharky87
2009-11-06 08:46:03

I was confused about the fact that I seemed to be getting a $1/160$ , so here's an explanation for those of you who were as confused as azot and I:

I was getting my r's confused in Yosun's solution, I just took $r=R$ and canceled away. The reason this is wrong is that the radius that you are taking the area with when multiplying by $\vec{E}$ is actually $R/2$ . So if you set $r=R/2$ you should arrive at the solution.

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jw111
2008-11-05 15:43:14

in addition, if you set A=1, R=1, epsilon=1

it will be easier

$E*4pi*\frac{1}{2^2}=\int r^2*4pi*r^2 dr$

then
$E \frac{1}{2^2} = \frac{1}{5} \frac{1}{2^5}$

E= 1/40

pam d
2011-09-17 16:24:27

nice

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gliese876d
2008-10-11 20:56:52

hmm. I'm just wondering... I had the following formula memorized for similar situations:

E= $\frac{Qr}{4piepsilonR^3}$

Where little r is the radius of the point from the center and R is the total radius of the sphere
Now using this formula I figured Q would equal $\frac{4piR^3rho}{3}$

so I used this in the numerator for Q and also plugged in r= R/2 and you get

$\frac{AR^3}{6epsilon}$

which is *close* to the right answer, but not quite... where does this standard formula fall apart in reasoning for this particular situation?

flyboy621
2010-11-02 21:12:58

I think you are assuming constant charge density, which is not the case here.

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hisperati
2007-10-31 20:29:18

To find the charge we need to integrate a charge density, that is per unit volume. The answers correspond to multipliying by a surface area ( I am not talking about the left side of Guass's Law, where it is obvious surface area is correctly used). Why isn't the charge density multiplied by $\frac{4}{3} pi R^3$ ?

blah22
2008-03-14 18:01:46

Yea I'm confused too

blah22
2008-03-14 18:27:59

Ah I see, for anyone else confused by this in the future, it's just a method of getting the volume of a sphere through integration. Divide the sphere into lots of shells of thickness dr, each with surface area $4 \Pi r^2$ .

Thanks so much Yosun!

flyboy621
2010-11-02 21:14:34

You can't just multiply the charge density by the enclosed volume because the density is not constant. You have to integrate.

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azot
2007-10-23 06:04:01

the answer to this problem is wrong .
E shoud be equal to A*R^3/32*5

azot
2007-10-23 06:05:00

sorry I was wrong :) Everything is ok

kevinjay15
2007-10-23 18:24:01

ETS is never wrong.

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Does anybody have a problem by using Coulumn' law? I used $\vec{E}=\frac{1}{4\Pi\e0}\int_0^\R/2\frac{Ar^2}{r^2}r^2Sin\Theta dr d\Theta d\Phi$ . The answer is $\frac{AT^4}{24e0}$ . Can anybody explain what I am wrong?

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